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Magazine Chapter 12: Problem 39
Find the distance from the point to the plane. $$(2,-3,4), \quad x+2 y+2 z=13$$
Short Answer
Expert verified
The distance from the point to the plane is 3 units.
Step by step solution
01
Identify the Given Elements
We are given a point \((2, -3, 4)\) and a plane defined by the equation \(x + 2y + 2z = 13\). Our task is to find the perpendicular distance from the point to the plane.
02
Write the Point's Coordinates
The point \((2, -3, 4)\) gives us the coordinates \(x_1 = 2\), \(y_1 = -3\), and \(z_1 = 4\).
03
Extract Plane's Normal Vector
The plane equation is \(x + 2y + 2z = 13\). The coefficients of \(x, y,\) and \(z\) give us the normal vector \( \mathbf{n} = (1, 2, 2) \).
04
Use Distance Formula
The formula to find the distance \(d\) from a point \((x_1, y_1, z_1)\) to a plane \(ax + by + cz = d\) is:\[ d = \frac{|ax_1 + by_1 + cz_1 - d|}{\sqrt{a^2 + b^2 + c^2}} \]Substitute \(a = 1\), \(b = 2\), \(c = 2\), \(d = 13\), \(x_1 = 2\), \(y_1 = -3\), and \(z_1 = 4\).
05
Compute the Dot Product
Calculate the numerator of the formula:\[ |1 \cdot 2 + 2 \cdot (-3) + 2 \cdot 4 - 13| = |2 - 6 + 8 - 13| = |-9| = 9 \]
06
Compute the Magnitude of the Normal Vector
Calculate the denominator of the formula:\[ \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3 \]
07
Calculate the Distance
Substitute the results from Steps 5 and 6 into the formula:\[ d = \frac{9}{3} = 3 \]
08
Final Step: Conclusion
The distance from the point \((2, -3, 4)\) to the plane \(x + 2y + 2z = 13\) is 3 units.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Plane Equation
A plane equation is a mathematical description that represents a flat, two-dimensional surface extending infinitely in three-dimensional space. In its standard form, this equation is presented as \(ax + by + cz = d\), where:
- \(a, b, c\) are the coefficients representing the plane.
- \(x, y, z\) are the variables that indicate any point on the plane.
- \(d\) is the constant term of the plane equation.
Normal Vector
The normal vector is a significant concept in understanding the orientation of a plane. It is a vector that is perpendicular (orthogonal) to every point on the plane. If a plane equation is given as \(ax + by + cz = d\), the normal vector \(\mathbf{n}\) is represented as \((a, b, c)\).
- It indicates the direction that is "up" from the plane.
- In our example, the normal vector for the plane \(x + 2y + 2z = 13\) is \(\mathbf{n} = (1, 2, 2)\).
Distance Formula
The distance formula provides a method to measure how far a point is from a plane along the perpendicular direction. For a point \((x_1, y_1, z_1)\) and a plane \(ax + by + cz = d\), the distance \(d\) is calculated as:\[d = \frac{|ax_1 + by_1 + cz_1 - d|}{\sqrt{a^2 + b^2 + c^2}}\]Here's how it works:
- The numerator \(|ax_1 + by_1 + cz_1 - d|\) accounts for the algebraic distance.
- The denominator \(\sqrt{a^2 + b^2 + c^2}\) normalizes this value using the magnitude of the normal vector.
Coordinates
Coordinates are numerical values that represent a point's position in space. In three-dimensional geometry, a point has three coordinates usually denoted as \((x, y, z)\). These values specify the location concerning three perpendicular axes鈥攃ommonly labeled as the x-axis, y-axis, and z-axis.
- For example, the coordinates of our point are \((2, -3, 4)\).
- This means the point is located 2 units along the x-axis, -3 units along the y-axis, and 4 units along the z-axis from the origin.
